Mixing
Alternating series remainder overestimation
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
$$sum^infty_n=0frac(-1)^n+1(2n+1)!$$ is approximated using the partial sum $$sum_n=0^3frac(-1)^n+1(2n+1)!$$
The error bound is $$frac(-1)^59!approx -0.000003$$
Why is this an overestimation and not an underestimation since it is negative?
calculus summation
asked Jul 27 at 0:29
Jinzu
328311
add a comment |Â
up vote
0
down vote
favorite
$$sum^infty_n=0frac(-1)^n+1(2n+1)!$$ is approximated using the partial sum $$sum_n=0^3frac(-1)^n+1(2n+1)!$$
The error bound is $$frac(-1)^59!approx -0.000003$$
Why is this an overestimation and not an underestimation since it is negative?
calculus summation
asked Jul 27 at 0:29
Jinzu
328311
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$sum^infty_n=0frac(-1)^n+1(2n+1)!$$ is approximated using the partial sum $$sum_n=0^3frac(-1)^n+1(2n+1)!$$
The error bound is $$frac(-1)^59!approx -0.000003$$
Why is this an overestimation and not an underestimation since it is negative?
calculus summation
asked Jul 27 at 0:29
Jinzu
328311
$$sum^infty_n=0frac(-1)^n+1(2n+1)!$$ is approximated using the partial sum $$sum_n=0^3frac(-1)^n+1(2n+1)!$$
The error bound is $$frac(-1)^59!approx -0.000003$$
Why is this an overestimation and not an underestimation since it is negative?
calculus summation
asked Jul 27 at 0:29
Jinzu
328311
asked Jul 27 at 0:29
Jinzu
328311
asked Jul 27 at 0:29
Jinzu
328311
asked Jul 27 at 0:29
Jinzu
328311
328311
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
The true value is between the partial sum and the partial sum plus the error bound. So, if the error is negative, the partial sum is an overestimate.
answered Jul 27 at 0:49
saulspatz
10.4k21323
So do you subtract the error bound from the partial sum?
– Jinzu
Jul 27 at 1:00
Exactly. If the partial sum is $s$, we know the sum of the series is between $s$ and $s-.000003,$ so $s$ overestimates the sum.
– saulspatz
Jul 27 at 1:02
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The true value is between the partial sum and the partial sum plus the error bound. So, if the error is negative, the partial sum is an overestimate.
answered Jul 27 at 0:49
saulspatz
10.4k21323
So do you subtract the error bound from the partial sum?
– Jinzu
Jul 27 at 1:00
Exactly. If the partial sum is $s$, we know the sum of the series is between $s$ and $s-.000003,$ so $s$ overestimates the sum.
– saulspatz
Jul 27 at 1:02
add a comment |Â
up vote
0
down vote
accepted
The true value is between the partial sum and the partial sum plus the error bound. So, if the error is negative, the partial sum is an overestimate.
answered Jul 27 at 0:49
saulspatz
10.4k21323
So do you subtract the error bound from the partial sum?
– Jinzu
Jul 27 at 1:00
Exactly. If the partial sum is $s$, we know the sum of the series is between $s$ and $s-.000003,$ so $s$ overestimates the sum.
– saulspatz
Jul 27 at 1:02
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The true value is between the partial sum and the partial sum plus the error bound. So, if the error is negative, the partial sum is an overestimate.
answered Jul 27 at 0:49
saulspatz
10.4k21323
The true value is between the partial sum and the partial sum plus the error bound. So, if the error is negative, the partial sum is an overestimate.
answered Jul 27 at 0:49
saulspatz
10.4k21323
answered Jul 27 at 0:49
saulspatz
10.4k21323
answered Jul 27 at 0:49
saulspatz
10.4k21323
answered Jul 27 at 0:49
saulspatz
10.4k21323
10.4k21323
So do you subtract the error bound from the partial sum?
– Jinzu
Jul 27 at 1:00
Exactly. If the partial sum is $s$, we know the sum of the series is between $s$ and $s-.000003,$ so $s$ overestimates the sum.
– saulspatz
Jul 27 at 1:02
add a comment |Â
So do you subtract the error bound from the partial sum?
– Jinzu
Jul 27 at 1:00
Exactly. If the partial sum is $s$, we know the sum of the series is between $s$ and $s-.000003,$ so $s$ overestimates the sum.
– saulspatz
Jul 27 at 1:02
So do you subtract the error bound from the partial sum?
– Jinzu
Jul 27 at 1:00
So do you subtract the error bound from the partial sum?
– Jinzu
Jul 27 at 1:00
Exactly. If the partial sum is $s$, we know the sum of the series is between $s$ and $s-.000003,$ so $s$ overestimates the sum.
– saulspatz
Jul 27 at 1:02
Exactly. If the partial sum is $s$, we know the sum of the series is between $s$ and $s-.000003,$ so $s$ overestimates the sum.
– saulspatz
Jul 27 at 1:02
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2863946%2falternating-series-remainder-overestimation%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password